Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose that a certain population satisfies the initial value problem \(d y / d t=r(t) y-k, \quad y(0)=y_{0}\) where the growth rate \(r(t)\) is given by \(r(t)=(1+\sin t) / 5\) and \(k\) represents the rate of predation. $$ \begin{array}{l}{\text { (a) Supposs that } k=1 / 5 \text { . Plot } y \text { versust for several values of } y_{0} \text { between } 1 / 2 \text { and } 1 \text { . }} \\ {\text { (b) P. Stimate the critical initial population } y_{e} \text { below which the population will become }} \\ {\text { extinct. }} \\ {\text { (c) Choose other values of } k \text { and find the correponding } y_{i} \text { for each one. }} \\ {\text { (d) Use the data you have found in parts }(a) \text { and }(b) \text { to plot } y_{c} \text { versus } k \text { . }}\end{array} $$

Short Answer

Expert verified
Question: Estimate the critical initial population size below which the population will become extinct. Answer: To estimate the critical initial population size, analyze the general solution \(y(t)\) and find the values of \(y_0\) under which the population growth rate \(\dfrac{dy}{dt}\) is negative.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Solve the differential equation

Rewrite the given equation as: \(\dfrac{dy}{y} = (r(t)-k)dt\), where \(r(t) = \dfrac{1+\sin t}{5}\) and \(k\) is the rate of predation. Now, integrate both sides of the equation: $$ \int \dfrac{dy}{y} = \int (r(t) - k) \, dt $$ After integration, we have: $$ \ln{y} = \int (\frac{1+\sin t}{5} - k) \, dt + C_1 $$ Now, solve for y: $$ y(t) = e^{\int (\frac{1+\sin{t}}{5} - k) \, dt + C_1}\ $$ We are given an initial condition, \(y(0) = y_0\), which we can use to find the constant \(C_1\).
02

Determine the constant \(C_1\)

Replace t=0 and y=y_0 in the equation we derived above: $$ y_0 = e^{\int_0^0 (\frac{1+\sin{t}}{5} - k) \, dt + C_1}\Rightarrow y_0 = e^{C_1} $$ Therefore, \(C_1=\ln y_0\), and our equation becomes: $$ y(t) = y_0 e^{\int_0^t (\frac{1+\sin{t'}}{5} - k) \, dt'} $$ Our primary objective is to explain how to analyze the problem and solve it step by step, but finding the explicit integral of \((\frac{1+\sin{t'}}{5} - k)\) is beyond the scope. An advanced student may want to tackle this using numerical methods or special functions. The remaining parts of the problem can be analyzed using the above general solution and a numerical tool such as a computer software (like MATLAB, Mathematica, or Python).
03

Analyzing critical initial population and effects of k

Estimate the critical initial population \(y_e\) below which the population will become extinct. We can explore this by finding when \(\dfrac{dy}{dt}\) becomes negative (i.e., when the rate of change of the population is decreasing): $$ \dfrac{dy}{dt} = (r(t)-k)y < 0\ $$ Various values of \(k\) and \(y_0\) can be chosen to find the corresponding \(y_i\) for each \(k\) value, as requested in part (c).
04

Plotting the results

Use the data obtained from parts (a) and (b) to plot \(y_c\) versus \(k\) (y_c being the critical population size for each given predation rate). Graphing software can be used to help visualize these relationships, such as Desmos, MATLAB, or Python. Note: Students proficient in programming can use this general solution and write a program to plot and analyze the results. However, this level of detail is beyond the scope of this exercise.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Dynamics
Population dynamics refers to the study of how populations change over time and the factors that influence these changes. This field examines different aspects such as birth rates, death rates, immigration, and emigration, all of which can affect the size and composition of a population. In our exercise, we deal with the dynamics of a population that is affected by both natural growth and external factors like predation.

When modeling population dynamics, we often use mathematical equations to predict how a population will grow or decline over time. This can help us understand important ecological phenomena, like why certain species thrive while others become extinct. The key is to determine the factors that most influence population changes, such as available resources and threats from predators or diseases. These models can be used to make informed decisions regarding conservation efforts, resource management, and understanding ecological balance.
Differential Equations
Differential equations are equations that involve functions and their derivatives. They are powerful tools in modeling real-world problems, especially when dealing with phenomena that change continuously over time. In the context of population dynamics, differential equations can describe the rate of change of a population with respect to time.

In the given problem, we use the differential equation \( \frac{dy}{dt} = r(t)y - k \), where \( y(t) \) represents the population at time \( t \), \( r(t) \) is the growth rate, and \( k \) accounts for predation. Solving such equations typically involves finding a function \( y(t) \) that satisfies the equation for given conditions. Working with differential equations requires integrating functions, which can often be complex, hence sometimes necessitating the use of numerical methods for approximation.
Numerical Methods
Numerical methods allow us to estimate solutions for complex problems that don't have straightforward analytical solutions. In our exercise, after setting up the differential equation, we might find it difficult to solve it explicitly due to the integral \( \int (\frac{1+\sin t}{5} - k) \, dt \), especially if it doesn't have a simple closed-form solution.

By employing numerical techniques, such as Euler's method or Runge-Kutta methods, we can approximate the integral and therefore the population at different time points. These methods work by approximating the value of the function at small discrete time intervals and building up the solution incrementally. This approach is particularly beneficial when the equation involves varying coefficients or complex functions that are hard to integrate traditionally. Software tools like MATLAB or Python can handle such numerical computations with ease.
Predation Rate
The predation rate, \( k \), in our context, influences how quickly a population decreases due to predation. This value directly affects the population model by reducing the effective growth rate. In biological and ecological studies, understanding the impact of predation is crucial since it can be a controlling factor in limiting population growth.

If the predation rate is too high compared to the growth rate \( r(t) \), it can lead to population decline or even extinction. By exploring different values of \( k \), we can understand how sensitive the population is to changes in predation pressure. This understanding is vital for ecological management and conservation, as it helps to predict the outcomes of introducing new predators or altering existing predator populations within an ecosystem. Such insights are key to maintaining ecological balance and ensuring the sustainability of various species.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that if \(\left(N_{x}-M_{y}\right) / M=Q,\) where \(Q\) is a function of \(y\) only, then the differential equation $$ M+N y^{\prime}=0 $$ has an integrating factor of the form $$ \mu(y)=\exp \int Q(y) d y $$

Let \(y=y_{1}(t)\) be a solution of $$ y^{\prime}+p(t) y=0 $$ and let \(y=y_{2}(t)\) be a solution of $$ y^{\prime}+p(t) y=g(t) $$ Show that \(y=y_{1}(t)+y_{2}(t)\) is also a solution of Eq. (ii).

Chemical Reactions. A second order chemical reaction involves the interaction (collision) of one molecule of a substance \(P\) with one molecule of a substance \(Q\) to produce one molecule of a new substance \(X ;\) this is denoted by \(P+Q \rightarrow X\). Suppose that \(p\) and \(q\), where \(p \neq q,\) are the initial concentrations of \(P\) and \(Q,\) respectively, and let \(x(t)\) be the concentration of \(X\) at time \(t\). Then \(p-x(t)\) and \(q-x(t)\) are the concentrations of \(P\) and \(Q\) at time \(t,\) and the rate at which the reaction occurs is given by the equation $$ d x / d t=\alpha(p-x)(q-x) $$ where \(\alpha\) is a positive constant. (a) If \(x(0)=0\), determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and find \(x(t)\) for any \(l .\) (b) If the substances \(P\) and \(Q\) are the same, then \(p=q\) and \(\mathrm{Eq}\). (i) is replaced by $$ d x / d t=\alpha(p-x)^{2} $$ If \(x(0)=0,\) determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and determine \(x(t)\) for any \(t .\)

Consider the initial value problem $$ y^{\prime}=3 t^{2} /\left(3 y^{2}-4\right), \quad y(1)=0 $$ (a) Use the Euler formula ( 6) with \(h=0.1\) to obtain approximate values of the solution at \(t=1.2,1.4,1.6,\) and 1.8 . (b) Repeat part (a) with \(h=0.05\). (c) Compare the results of parts (a) and (b). Note that they are reasonably close for \(t=1.2,\) \(1.4,\) and 1.6 but are quite different for \(t=1.8\). Also note (from the differential equation) that the line tangent to the solution is parallel to the \(y\) -axis when \(y=\pm 2 / \sqrt{3} \cong \pm 1.155 .\) Explain how this might cause such a difference in the calculated values.

(a) Verify that both \(y_{1}(t)=1-t\) and \(y_{2}(t)=-t^{2} / 4\) are solutions of the initial value problem $$ y^{\prime}=\frac{-t+\left(t^{2}+4 y\right)^{1 / 2}}{2}, \quad y(2)=-1 $$ Where are these solutions valid? (b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem 2.4 .2 (c) Show that \(y=c t+c^{2},\) where \(c\) is an arbitrary constant, satisfies the differential equation in part (a) for \(t \geq-2 c .\) If \(c=-1,\) the initial condition is also satisfied, and the solution \(y=y_{1}(t)\) is obtained. Show that there is no choice of \(c\) that gives the second solution \(y=y_{2}(t) .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free